Composite function problems in high school mathematics, high school mathematics, finding compound fu

Someone has already given the right answer. Let's talk about why your method is wrong.

f(x+1)=(x+1) 2+1, where x can take any value, then x+1 can also take any value, so x+1 can be replaced with a variable t that can take any value. That is, regardless of the value of the independent variable t, the function satisfies the relation f(t)=(t) 2+1. t and x+1 have the same range, so substitution makes sense.

But with f(f(x)-x 2+x)=f(x)-x 2+x, it doesn't make sense for you to replace f(x)-x 2+x with an arbitrary variable t. This is because f(x)-x 2+x may not be able to get any value. (It is conceivable that when f(x)=x, f(x)-x 2+x can only be taken as a negative value, and there is no positive value).

So the range of t and f(x)-x 2+x is not necessarily the same. Since they are different, they cannot be substituted for each other.

1: f(2)=3, let x=2

then f(f(2)-2 +2)=f(3-4+2)=f(1)=f(x)-x 2+x=f(2)-2 2+2=3-4+2=1

f(0)=a, so that x=0, then f(f(0)-0 2+0)=f(a)=f(x)-x 2+x=f(0)-0 2+0=a

2: These two questions are completely two types of questions, corresponding to the second question, the general practice is to make x+1=t, and then get f(t)=t 2+1, where the relationship between t and x is very clear, not a composite function.

According to this practice, let f(x)-x 2+x=t, where t and x contain f(x), which is a composite function and cannot be simply replaced directly.

i) Analysis: x r, there is f(f(x)-x 2+x) = f(x)-x 2+x, f(2)=3

f(f(2)-2 2+2)=f(2)-2 2+2=1 f(0)=a, then f(f(0)-0 2+0)=f(0)-0 2+0=a, i.e., f(a)=a

It is known that f(x+1)=(x+1) 2+1 is f(x).

Let t=x+1==>x=t-1

f(t)=(t-1+1)^2+1=t^2+1

f(x)=x^2+1

It can be seen that the result of f(x) is not the result of direct substitution, although, as with the result of direct substitution, this is only a coincidence and not a universal law.

For example, f(x+1)=x 2+1 is known to find f(x).

Let t=x+1==>x=t-1

f(t)=(t-1)^2+1=t^2-2t+2

f(x)=x^2-2x+1

Let's find f(x).

Analysis: Suppose f(x)=x

x r, there is f(f(x)-x 2+x)=f(x)-x 2+x

f(x)-x^2+x=x==>f(x)=x^2

Obviously not in line with the title.

Suppose there is one and only one real number, such that f(x0)=x0

For any x r, there is f(x)-x 2+x =x0

Let x=x0, there is f(x0)-x02+x0=x0

f(x0)=x0, so x0-x0 2=0, so x0=0 or x0=1

If x0=0, then f(x)-x 2+x=0, i.e. f(x)=x 2-x

The equation x 2-x=x has two different real roots, which contradict the condition of the problem, so x0≠0

If x0=1, then there is f(x)-x 2+x=1, that is, f(x)=x 2-x+1 This function satisfies the question.

In summary, the calculated function is f(x)=x 2-x+1(x r).

Your solution is not correct, and the other problem can be directly replaced with x=x-1 - which means assignment.

The f(x) range should be discussed on a case-by-case basis, whether it is greater than 0 or less than 0. Then apply f(x) to the corresponding function. See**.

Known step function: f(x)=1+x, (x<0)...f(x)=x，(x≧0)..Find f[f(x)]=

Solution: f[f(x)]=1+f(x)=1+1+x=2+x, when f(x)=1+x<0, i.e. x<-1];

.From: f[f(x)]=f(x)=x, [when f(x)=x 0, i.e. x 0];

If you see this message remember to reply, I'll be there.

Personally, I think this question is a bit problematic, and the question should be changed to have two unequal real roots, but according to the process and the answer you gave, it is derived from a comprehensive analysis. Here I will talk to you according to my thoughts, if you have any questions, please ask.

The image is drawn first according to the function f(x). Sorting out (2-x) x=(2 x)-1 (dividing the numerator by the denominator), we know that when x>=1, it is an inverse proportional function about x, which is obtained by moving y=2 x down by one unit, infinitely approaching y=-1. But x = 1, so the image is part of it.

Draw an image of 2 (x-1), which is an exponential function, which is derived from the function y=2 x shifted one unit to the right. Take the part of image x<1.

In this way, the image is drawn like a "person" with a (1,1) top

From g(x)=x 2-2x, the minimum value of g(x) is -1 (when x=1). So the minimum x in f(x) starts at -1. Draw the line y=k (a straight line parallel to the y-axis), because there are two unequal real roots, so there should be two intersections between the straight line y=k and the f(x) image.

From the image, it can be seen that when x=-1, f(x)=1 4, when y=1 4, there are exactly two intersection points, and it is okay to translate y=k upward, but when it reaches the vertex (1,1), there is only one intersection point, which cannot be taken, so k belongs to (1 4,1).

This is my way of thinking, and what the teacher said should be roughly the same, and this is the way of thinking. As for the 4 real roots given in the question, I really don't think it's appropriate, this is a good question.

Please also listen carefully to the teacher's analysis in class, and if the question is correct, please tell me what I did wrong and what should I do. Thank you.

Big brother, it's really hard to say, 1 is the maximum value of the composite function, and the quarter is the local minimum, you have to draw the function image first.

The function y=f(u)=u, not like we call it the parent function, its definition domain d=[0,+ function u=g(x)=tanx, not like we call it a subfunction, its definition domain d=if these two functions can form a composite function f(g(x))=tanx), then you must modify the definition domain of the subfunction u=g(x)=tanx, that is, the subfunction should keep its original definition d= cannot form a composite function with the parent function, why?

If the composite function f(g(x))=tanx) holds.

y=f(u)= u, which defines the domain d=[0,+ u[0,+

Again, the parent function definition domain is the value range of the child function.

u=g(x)=tanx [0,+ The domain of the + subfunction is k <=xk - 2< x

See, that is, in order for two functions to form a composite function, the value range of one function must be satisfied, and the range of the definition domain of the other function is within the range of the definition domain of the other function, and the definition domain of the y function is d, and the value range of the g function is infinite, so the condition is not satisfied, and in order to satisfy the condition, the value range of g must be taken from the definition domain of the y function.

The one below the red line is the one that controls the independent variable of the g function so that its range is within [0 positive infinity), and then its range is in the defined domain d of the y function.

You can draw a graph of the function of tanx.

d means that the value range of x is in the first three quadrants, because tanx is greater than 0 in the first three quadrants, so it can be squared, so it can form a coincidence function. The condition for conforming to a function is that the range of values of the inner function is contained in the defined domain of the outer function.

f(x) is x instead of x in f(x).

So f(x)=2x -1

In the same way, f]g(x)] is to replace xf[g(x)]=2 (x +1)-1 in f(x) with 1 x +1

g[f(x)+2]

f(2x+1)

1/[(2x+1)²+1]

1/(4x²+4x+2)

f(2x+1)=x²-2x

Requirement f(2).

Then let 2x+1=2

x=1 2So(2)=(1 2) -2 1 2=-3 4f(x)+2f(1 x)=3x

Then let x=2f(2)+2f(1 2)=6 (1).

Let x=1 2

f(1/2)+2f(2)=3/2 (2)

3f(2)=-3

f(2)=-1

2f(x)+f(-x)=3x+2

Let x=2 then 2f(2)+f(-2)=3 2+2

f(-2)=-16/3

So f(2)=20 3

Requirement f(2).

So the key to these questions is to make 2 in parentheses after f, and if you find f(x).

Then x and 1 x and x and -x can be substituted in the last two.

For example, let x=1 x, and get a new functional equation, which can be solved with the original coupling.

The first one directly substitutes x for half to get f(2), and the second one replaces x for 1 x, remember, and the x in the next 3x also needs to be changed! Then subtract the two formulas to get the analytic formula of the function! Substitute to get the answer!

For the third one, just replace x with 2!

To do this kind of question, the main thing is to know the overall substitution! After changing, there is no way to form the original form! But some can be opportunistic! Such as the first one! Haha, it's actually better to practice more! Think more! Finally, good luck!

f(x) is x instead of x in f(x), so f(x)=2x -1 In the same way, f]g(x)] is to replace x in f(x) with 1 x 1 f[g(x)]=2 (x 1)-1 g[f(x) 2].

Solution: From the value range of the function f(x)=log (x -ax-a) is r, the function y=x -ax-a passes through all positive numbers, so there is =a +4a 0, and a -4 or a 0 is obtained, and then according to f(x) on (-3,1-3) is the increasing function, the function y=x -ax-a is a subtraction function on (-3,1-3) and is a positive value, so a 2 1-3, and when x=1-3, y>0, that is, a 2-2 3, and 4-2 3-a(1- 3)-a>0, find 2-2 3 a<2, 0 a<2.

2°, <0 gets -4

(1) Composite function y=f(u),u=g(x)When it is not compounded, the domain of the function y=f(u) is the range of values of u, and after compounding, the domain of the function y=f[g(x)] is the range of values of x. However, the value range of the inner function should not exceed the defined domain of the outer function.

My approach to this question is, 0 x 1,===>-1 x-1 0.This finds the domain of the function y=f(x) as [-1,0].∴1≤x+2≤0.

=>-3≤x≤-2.The domain of the function f(x+2) is [-3,-2].(2) f(x+2)=1 f(x)

=>f(x+4)=1/f(x+2)=f(x).The function f(x) is a periodic function with period 4. f(5)=f(1)=-5.

As can be seen from the foregoing, f(-3)=1 f(-5)f(1)=f(-3).∴f(-5)=1/f(-3)=1/f(1)=-1/5.

f[f(5)]=f(-5)=-1/5.

1.x-1 and x+2 have the same value range, and the definition field refers to the range of x, so x is different. 2.

Medium can know that f(x) is a periodic function. In the same equation, the same unknown must be the same. I don't know if you understand?

The domain of f(x-1) refers to the range of x, and the two x's are not the same.